If int frac{cos 3x}{sin x} dx = p cos 2x + q | vedclass

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(B) We are given the integral $I = \int \frac{\cos 3x}{\sin x} dx$. Using the trigonometric identity $\cos 3x = 4 \cos^3 x - 3 \cos x$,we can also wri

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$2$

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(B) We are given the integral $I = \int \frac{\cos 3x}{\sin x} dx$. Using the trigonometric identity $\cos 3x = 4 \cos^3 x - 3 \cos x$,we can also write $\cos 3x = \cos(2x + x) = \cos 2x \cos x - \sin 2x \sin x$. Alternatively,use $\cos 3x = 1 - 4 \sin^2 x$ is incorrect,rather $\cos 3x = \cos(2x+x) = \cos 2x \cos x - \sin 2x \sin x = (1 - 2 \sin^2 x) \cos x - (2 \sin x \cos x) \sin x = \cos x - 2 \sin^2 x \cos x - 2 \sin^2 x \cos x = \cos x - 4 \sin^2 x \cos x$. Thus,$\frac{\cos 3x}{\sin x} = \frac{\cos x - 4 \sin^2 x \cos x}{\sin x} = \cot x - 4 \sin x \cos x = \cot x - 2 \sin 2x$. Now,integrate: $\int (\cot x - 2 \sin 2x) dx = \int \cot x dx - 2 \int \sin 2x dx$. $= \log |\sin x| - 2 (-\frac{\cos 2x}{2}) + C = \log |\sin x| + \cos 2x + C$. Comparing this with $p \cos 2x + q \log |\sin x| + C$,we get $p = 1$ and $q = 1$. Therefore,$p + q = 1 + 1 = 2$.

If $\int \frac{\cos 3x}{\sin x} dx = p \cos 2x + q \log |\sin x| + C$,then $p + q =$ . . . . . . .

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